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Multivariate Linear Models in R*

An Appendix to An R Companion to Applied Regression, third edition

John Fox & Sanford Weisberg

last revision: 2018-09-21

Abstract

The multivariate linear model is

Y(n×m)

= X(n×k+1)

B(k+1×m)

+ E(n×m)

where Y is a matrix of n cases on m response variables; X is a model matrix with columnsfor k + 1 regressors, typically including an initial column of 1s for the regression constant; Bis a matrix of regression coefficients, one column for each response variable; and E is a matrixof errors. This model can be fit with the lm() function in R, where the left-hand side of themodel comprises a matrix of response variables, and the right-hand side is specified exactly asfor a univariate linear model (i.e., with a single response variable). This appendix to Fox andWeisberg (2019) explains how to use the Anova() and linearHypothesis() functions in the carpackage to test hypotheses for parameters in multivariate linear models, including models forrepeated-measures data.

1 Basic Ideas

The multivariate linear model accommodates two or more response variables. The theory of multi-variate linear models is developed very briefly in this section. Much more extensive treatments maybe found in the recommended reading for this appendix.

The multivariate general linear model is

Y(n×m)

= X(n×k+1)

B(k+1×m)

+ E(n×m)

where Y is a matrix of n cases on m response variables; X is a model matrix with columns for k+ 1regressors, typically including an initial column of 1s for the regression constant; B is a matrix ofregression coefficients, one column for each response variable; and E is a matrix of errors.1 Thecontents of the model matrix are exactly as in the univariate linear model (as described in Chapter 4of An R Companion to Applied Regression, Fox and Weisberg, 2019—hereafter, the“R Companion”),and may contain, therefore, dummy regressors representing factors, polynomial or regression-splineterms, interaction regressors, and so on.

The assumptions of the multivariate linear model concern the behavior of the errors: Let ε′irepresent the ith row of E. Then ε′i ∼ Nm(0,Σ), where Σ is a nonsingular error-covariance matrix,constant across cases; ε′i and ε′i′ are independent for i 6= i′; and X is fixed or independent of E.We can write more compactly that vec(E) ∼ Nnm(0, In ⊗Σ). Here, vec(E) ravels the error matrixrow-wise into a vector, and ⊗ is the Kronecker-product operator.

1A typographical note: B and E are, respectively, the upper-case Greek letters Beta and Epsilon. Because these areindistinguishable from the corresponding Roman letters B and E, we will denote the estimated regression coefficientsas B and the residuals as E.

1

The maximum-likelihood estimator of B in the multivariate linear model is equivalent to equation-by-equation least squares for the individual responses:

B = (X′X)−1X′Y

Procedures for statistical inference in the multivariate linear model, however, take account of thefact that there are several, generally correlated, responses.

Paralleling the decomposition of the total sum of squares into regression and residual sums ofsquares in the univariate linear model, there is in the multivariate linear model a decomposition ofthe total sum-of-squares-and-cross-products (SSP) matrix into regression and residual SSP matrices.We have

SSPT(m×m)

= Y′Y − ny y′

= E′E +(Y′Y − ny y′

)= SSPR + SSPReg

where y is the (m × 1) vector of means for the response variables; Y = XB is the matrix of fitted

values; and E = Y − Y is the matrix of residuals.Many hypothesis tests of interest can be formulated by taking differences in SSPReg (or, equiva-

lently, SSPR) for nested models. Let SSPH represent the incremental SSP matrix for a hypothesis.Multivariate tests for the hypothesis are based on the m eigenvalues λj of SSPHSSP−1R (the hy-pothesis SSP matrix “divided by” the residual SSP matrix), that is, the values of λ for which

det(SSPHSSP−1R − λIm) = 0

The several commonly employed multivariate test statistics are functions of these eigenvalues:

Pillai-Bartlett Trace, TPB =

m∑j=1

λj1− λj

Hotelling-Lawley Trace, THL =

m∑j=1

λj

Wilks’s Lambda, Λ =

m∏j=1

1

1 + λj

Roy’s Maximum Root, λ1

(1)

By convention, the eigenvalues of SSPHSSP−1R are arranged in descending order, and so λ1 is thelargest eigenvalue. There are F approximations to the null distributions of these test statistics. Forexample, for Wilks’s Lambda, let s represent the degrees of freedom for the term that we are testing(i.e., the number of columns of the model matrix X pertaining to the term). Define

r = n− k − 1− m− s+ 1

2(2)

u =ms− 2

4

t =

√m2s2 − 4

m2 + s2 − 5for m2 + s2 − 5 > 0

0 otherwise

2

Rao (1973, p. 556) shows that under the null hypothesis,

F0 =1− Λ1/t

Λ1/t× rt− 2u

ms(3)

follows an approximate F -distribution with ms and rt− 2u degrees of freedom, and that this resultis exact if min(m, s) ≤ 2 (a circumstance under which all four test statistics are equivalent).

Even more generally, suppose that we want to test the linear hypothesis

H0: L(q×k+1)

B(k+1×m)

= C(q×m)

(4)

where L is a hypothesis matrix of full-row rank q ≤ k+ 1, and the right-hand-side matrix C consistsof constants (usually 0s).2 Then the SSP matrix for the hypothesis is

SSPH =(B′L′ −C′

) [L(X

′X)−1

L′]−1 (

LB−C)

and the various test statistics are based on the p = min(q,m) nonzero eigenvalues of SSPHSSP−1R

(and the formulas in Equations 1, 2, and 3 are adjusted by substituting p for m).When a multivariate response arises because a variable is measured on different occasions, or

under different circumstances (but for the same individuals), it is also of interest to formulatehypotheses concerning comparisons among the responses. This situation, called a repeated-measuresdesign, can be handled by linearly transforming the responses using a suitable model matrix, forexample extending the linear hypothesis in Equation 4 to

H0: L(q×k+1)

B(k+1×m)

P(m×v)

= C(q×v)

(5)

Here, the response-transformation matrix P provides contrasts in the responses (see, e.g., Hand andTaylor, 1987, or O’Brien and Kaiser, 1985). The SSP matrix for the hypothesis is

SSPH(q×q)

=(P′B′L′ −C′

) [L(X

′X)−1

L′]−1 (

LBP−C)

and test statistics are based on the p = min(q, v) nonzero eigenvalues ofSSPH(P′SSPRP)−1.

2 Fitting and Testing Multivariate Linear Models in R

Multivariate linear models are fit in R with the lm() function. The procedure is the essence ofsimplicity: The left-hand side of the model is a matrix of responses, with each column representinga response variable and each row a case; the right-hand side of the model and all other argumentsto lm are precisely the same as for a univariate linear model (as described in Chapter 4 of the RCompanion). Typically, the response matrix is composed from individual response variables via thecbind() function.

The anova() function in the standard R distribution is capable of handling multivariate linearmodels (see Dalgaard, 2007), but the Anova() and linearHypothesis() functions in the car pack-age may also be employed, in a manner entirely analogous to that described in the R Companion(Section 5.3) for univariate linear models. We briefly demonstrate the use of these functions in thissection.

To illustrate multivariate linear models, we will use data collected by Anderson (1935) on threespecies of irises in the Gaspe Peninsula of Quebec, Canada. The data are of historical interest instatistics, because they were employed by R. A. Fisher (1936) to introduce the method of discriminantanalysis. The data frame iris is part of the standard R distribution:

2Cf., Section 5.3.5 of the R Companion for linear hypotheses in univariate linear models.

3

Figure 1: Three species of irises in the Anderson/Fisher data set: setosa (left), versicolor (center),and virginica (right). Source: The photographs are respectively by Radomil Binek, Danielle Langlois, and Frank

Mayfield, and are distributed under the Creative Commons Attribution-Share Alike 3.0 Unported license (first and

second images) or 2.0 Creative Commons Attribution-Share Alike Generic license (third image); they were obtained

from the Wikimedia Commons.

library(car)

Loading required package: carData

some(iris)

Sepal.Length Sepal.Width Petal.Length Petal.Width Species

25 4.8 3.4 1.9 0.2 setosa

47 5.1 3.8 1.6 0.2 setosa

67 5.6 3.0 4.5 1.5 versicolor

73 6.3 2.5 4.9 1.5 versicolor

104 6.3 2.9 5.6 1.8 virginica

109 6.7 2.5 5.8 1.8 virginica

113 6.8 3.0 5.5 2.1 virginica

131 7.4 2.8 6.1 1.9 virginica

140 6.9 3.1 5.4 2.1 virginica

149 6.2 3.4 5.4 2.3 virginica

The first four variables in the data set represent measurements (in cm) of parts of the flowers, whilethe final variable specifies the species of iris. (Sepals are the green leaves that comprise the calyxof the plant, which encloses the flower.) Photographs of examples of the three species of irises—setosa, versicolor, and virginica—appear in Figure 1. Figure 2 is a scatterplot matrix of the fourmeasurements classified by species, showing within-species 50 and 95% concentration ellipses (seeSection 5.2.3 of the R Companion); Figure 3 shows boxplots for each of the responses by species:

scatterplotMatrix(~ Sepal.Length + Sepal.Width + Petal.Length

+ Petal.Width | Species,

data=iris, smooth=FALSE, regLine=FALSE, ellipse=TRUE,

by.groups=TRUE, diagonal=FALSE, legend=list(coords="bottomleft"))

par(mfrow=c(2, 2))

for (response in c("Sepal.Length", "Sepal.Width",

"Petal.Length", "Petal.Width"))

Boxplot(iris[, response] ~ Species, data=iris, ylab=response)

As the photographs suggest, the scatterplot matrix and boxplots for the measurements reveal thatversicolor and virginica are more similar to each other than either is to setosa. Further, the ellipses

4

setosaversicolorvirginica

Sepal.Length

2.0

2.5

3.0

3.5

4.0

4.5 5.5 6.5 7.5

0.5

1.0

1.5

2.0

2.5

2.0 2.5 3.0 3.5 4.0

Sepal.Width

Petal.Length

1 2 3 4 5 6 7

0.5 1.0 1.5 2.0 2.5

4.5

5.5

6.5

7.5

12

34

56

7Petal.Width

Figure 2: Scatterplot matrix for the Anderson/Fisher iris data, showing within-species 50 and 95%concentration ellipses.

5

setosa versicolor virginica

4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

Species

Sep

al.L

engt

h

107


2.0

2.5

3.0

3.5

4.0

Species

Sep

al.W

idth

42


12

34

56

7

Species

Pet

al.L

engt

h

23

99


0.5

1.0

1.5

2.0

2.5

Species

Pet

al.W

idth

2444

Figure 3: Boxplots for the response variables in the iris data set classified by species.

6

in the scatterplot matrix suggest that the assumption of constant within-group covariance matricesis problematic: While the shapes and sizes of the concentration ellipses for versicolor and virginicaare reasonably similar, the shapes and sizes of the ellipses for setosa are different from the other two.

We proceed nevertheless to fit a multivariate one-way ANOVA model to the iris data:

mod.iris <- lm(cbind(Sepal.Length, Sepal.Width, Petal.Length, Petal.Width)

~ Species, data=iris)

class(mod.iris)

[1] "mlm" "lm"

mod.iris

Call:

lm(formula = cbind(Sepal.Length, Sepal.Width, Petal.Length, Petal.Width) ~

Species, data = iris)

Coefficients:

Sepal.Length Sepal.Width Petal.Length Petal.Width

(Intercept) 5.006 3.428 1.462 0.246

Speciesversicolor 0.930 -0.658 2.798 1.080

Speciesvirginica 1.582 -0.454 4.090 1.780

summary(mod.iris)

Response Sepal.Length :

Call:

lm(formula = Sepal.Length ~ Species, data = iris)

Residuals:

Min 1Q Median 3Q Max

-1.688 -0.329 -0.006 0.312 1.312

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 5.0060 0.0728 68.76 < 2e-16

Speciesversicolor 0.9300 0.1030 9.03 8.8e-16

Speciesvirginica 1.5820 0.1030 15.37 < 2e-16

Residual standard error: 0.515 on 147 degrees of freedom

Multiple R-squared: 0.619, Adjusted R-squared: 0.614

F-statistic: 119 on 2 and 147 DF, p-value: <2e-16

Response Sepal.Width :

Call:

lm(formula = Sepal.Width ~ Species, data = iris)

Residuals:


7

-1.128 -0.228 0.026 0.226 0.972

Coefficients:


(Intercept) 3.4280 0.0480 71.36 < 2e-16

Speciesversicolor -0.6580 0.0679 -9.69 < 2e-16

Speciesvirginica -0.4540 0.0679 -6.68 4.5e-10



F-statistic: 49.2 on 2 and 147 DF, p-value: <2e-16

Response Petal.Length :

Call:

lm(formula = Petal.Length ~ Species, data = iris)

Residuals:


-1.260 -0.258 0.038 0.240 1.348

Coefficients:


(Intercept) 1.4620 0.0609 24.0 <2e-16

Speciesversicolor 2.7980 0.0861 32.5 <2e-16

Speciesvirginica 4.0900 0.0861 47.5 <2e-16



F-statistic: 1.18e+03 on 2 and 147 DF, p-value: <2e-16

Response Petal.Width :

Call:

lm(formula = Petal.Width ~ Species, data = iris)

Residuals:


-0.626 -0.126 -0.026 0.154 0.474

Coefficients:


(Intercept) 0.2460 0.0289 8.5 2e-14

Speciesversicolor 1.0800 0.0409 26.4 <2e-16

Speciesvirginica 1.7800 0.0409 43.5 <2e-16



F-statistic: 960 on 2 and 147 DF, p-value: <2e-16

8

The lm() function returns an S3 object of class c("mlm", "lm"). The printed representation of theobject simply shows the estimated regression coefficients for each response, and the model summary isthe same as we would obtain by performing separate least-squares regressions for the four responses.

We use the Anova() function in the car package to test the null hypothesis that the four responsemeans are identical across the three species of irises:3

(manova.iris <- Anova(mod.iris))

Type II MANOVA Tests: Pillai test statistic

Df test stat approx F num Df den Df Pr(>F)

Species 2 1.19 53.5 8 290 <2e-16

class(manova.iris)

[1] "Anova.mlm"

summary(manova.iris)

Type II MANOVA Tests:

Sum of squares and products for error:


Sepal.Length 38.956 13.6300 24.6246 5.6450

Sepal.Width 13.630 16.9620 8.1208 4.8084

Petal.Length 24.625 8.1208 27.2226 6.2718

Petal.Width 5.645 4.8084 6.2718 6.1566

------------------------------------------

Term: Species

Sum of squares and products for the hypothesis:


Sepal.Length 63.212 -19.953 165.25 71.279

Sepal.Width -19.953 11.345 -57.24 -22.933

Petal.Length 165.248 -57.240 437.10 186.774

Petal.Width 71.279 -22.933 186.77 80.413

Multivariate Tests: Species


Pillai 2 1.192 53.47 8 290 < 2.2e-16

Wilks 2 0.023 199.15 8 288 < 2.2e-16

Hotelling-Lawley 2 32.477 580.53 8 286 < 2.2e-16

Roy 2 32.192 1166.96 4 145 < 2.2e-16

The Anova() function returns an object of class "Anova.mlm" which, when printed, produces amultivariate-analysis-of-variance (“MANOVA”) table, by default reporting Pillai’s test statistic; sum-marizing the object produces a more complete report. The object returned by Anova() may alsobe used in further computations, for example, for displays such as HE plots (Friendly, 2007; Fox

3The Manova() function in the car package is equivalent to Anova() applied to a multivariate linear model.

9

et al., 2009; Friendly, 2010). Because there is only one term (beyond the regression constant) on theright-hand side of the model, in this example the type-II test produced by default by Anova() is thesame as the sequential test produced by the standard R anova() function:

anova(mod.iris)

Analysis of Variance Table

Df Pillai approx F num Df den Df Pr(>F)

(Intercept) 1 0.993 5204 4 144 <2e-16

Species 2 1.192 53 8 290 <2e-16

Residuals 147

The null hypothesis is soundly rejected.The linearHypothesis() function in the car package may be used to test more specific hy-

potheses about the parameters in the multivariate linear model. For example, to test for differencesbetween setosa and the average of versicolor and virginica, and for differences between versicolorand virginica:

linearHypothesis(mod.iris, "0.5*Speciesversicolor + 0.5*Speciesvirginica",

verbose=TRUE)

Hypothesis matrix:

(Intercept) Speciesversicolor

0.5*Speciesversicolor + 0.5*Speciesvirginica 0 0.5

Speciesvirginica

0.5*Speciesversicolor + 0.5*Speciesvirginica 0.5

Right-hand-side matrix:

Sepal.Length Sepal.Width Petal.Length

0.5*Speciesversicolor + 0.5*Speciesvirginica 0 0 0

Petal.Width

0.5*Speciesversicolor + 0.5*Speciesvirginica 0

Estimated linear function (hypothesis.matrix %*% coef - rhs):


1.256 -0.556 3.444 1.430



Sepal.Length 52.585 -23.278 144.189 59.869

Sepal.Width -23.278 10.305 -63.829 -26.503

Petal.Length 144.189 -63.829 395.371 164.164

Petal.Width 59.869 -26.503 164.164 68.163



Sepal.Length 38.956 13.6300 24.6246 5.6450

Sepal.Width 13.630 16.9620 8.1208 4.8084

Petal.Length 24.625 8.1208 27.2226 6.2718

Petal.Width 5.645 4.8084 6.2718 6.1566

10

Multivariate Tests:


Pillai 1 0.9673 1063.9 4 144 < 2.2e-16

Wilks 1 0.0327 1063.9 4 144 < 2.2e-16


Roy 1 29.5520 1063.9 4 144 < 2.2e-16

linearHypothesis(mod.iris, "Speciesversicolor = Speciesvirginica",

verbose=TRUE)

Hypothesis matrix:

(Intercept) Speciesversicolor Speciesvirginica

Speciesversicolor = Speciesvirginica 0 1 -1

Right-hand-side matrix:


Speciesversicolor = Speciesvirginica 0 0 0

Petal.Width

Speciesversicolor = Speciesvirginica 0

Estimated linear function (hypothesis.matrix %*% coef - rhs):


-0.652 -0.204 -1.292 -0.700



Sepal.Length 10.6276 3.3252 21.0596 11.41

Sepal.Width 3.3252 1.0404 6.5892 3.57

Petal.Length 21.0596 6.5892 41.7316 22.61

Petal.Width 11.4100 3.5700 22.6100 12.25



Sepal.Length 38.956 13.6300 24.6246 5.6450

Sepal.Width 13.630 16.9620 8.1208 4.8084

Petal.Length 24.625 8.1208 27.2226 6.2718

Petal.Width 5.645 4.8084 6.2718 6.1566

Multivariate Tests:


Pillai 1 0.74525 105.31 4 144 < 2.2e-16

Wilks 1 0.25475 105.31 4 144 < 2.2e-16


Roy 1 2.92535 105.31 4 144 < 2.2e-16

The argument verbose=TRUE to linearHypothesis() shows the hypothesis matrix L and right-hand-side matrix C for the linear hypothesis in Equation 4 (page 3). In this case, all of the multi-variate test statistics are equivalent and therefore translate into identical F -statistics. Both focussednull hypotheses are easily rejected, but the evidence for differences between setosa and the other

11

two iris species is much stronger than for differences between versicolor and virginica. Testing that"0.5*Speciesversicolor + 0.5*Speciesvirginica" is 0 tests that the average of the mean vec-tors for these two species is equal to the mean vector for setosa, because the latter is the baselineccategory for the Species dummy regressors.

An alternative, equivalent, and in a sense more direct approach is to fit the model with customcontrasts for the three species of irises, followed up by a test for each contrast:

C <- matrix(c(1, -0.5, -0.5, 0, 1, -1), 3, 2)

colnames(C) <- c("setosa vs. versicolor & virginica", "versicolor & virginica")

contrasts(iris$Species) <- C

contrasts(iris$Species)

setosa vs. versicolor & virginica versicolor & virginica

setosa 1.0 0

versicolor -0.5 1

virginica -0.5 -1

(mod.iris.2 <- update(mod.iris))

Call:

lm(formula = cbind(Sepal.Length, Sepal.Width, Petal.Length, Petal.Width) ~

Species, data = iris)

Coefficients:


(Intercept) 5.843 3.057 3.758

Speciessetosa vs. versicolor & virginica -0.837 0.371 -2.296

Speciesversicolor & virginica -0.326 -0.102 -0.646

Petal.Width

(Intercept) 1.199

Speciessetosa vs. versicolor & virginica -0.953

Speciesversicolor & virginica -0.350

linearHypothesis(mod.iris.2, c(0, 1, 0)) # setosa vs. versicolor & virginica



Sepal.Length 52.585 -23.278 144.189 59.869

Sepal.Width -23.278 10.305 -63.829 -26.503

Petal.Length 144.189 -63.829 395.371 164.164

Petal.Width 59.869 -26.503 164.164 68.163



Sepal.Length 38.956 13.6300 24.6246 5.6450

Sepal.Width 13.630 16.9620 8.1208 4.8084

Petal.Length 24.625 8.1208 27.2226 6.2718

Petal.Width 5.645 4.8084 6.2718 6.1566

Multivariate Tests:


12

Pillai 1 0.9673 1063.9 4 144 < 2.2e-16

Wilks 1 0.0327 1063.9 4 144 < 2.2e-16


Roy 1 29.5520 1063.9 4 144 < 2.2e-16

linearHypothesis(mod.iris.2, c(0, 0, 1)) # versicolor vs. virginica



Sepal.Length 10.6276 3.3252 21.0596 11.41

Sepal.Width 3.3252 1.0404 6.5892 3.57

Petal.Length 21.0596 6.5892 41.7316 22.61

Petal.Width 11.4100 3.5700 22.6100 12.25



Sepal.Length 38.956 13.6300 24.6246 5.6450

Sepal.Width 13.630 16.9620 8.1208 4.8084

Petal.Length 24.625 8.1208 27.2226 6.2718

Petal.Width 5.645 4.8084 6.2718 6.1566

Multivariate Tests:


Pillai 1 0.74525 105.31 4 144 < 2.2e-16

Wilks 1 0.25475 105.31 4 144 < 2.2e-16


Roy 1 2.92535 105.31 4 144 < 2.2e-16

Finally, we can code the response-transformation matrix P in Equation 5 (page 3) to computelinear combinations of the responses, either via the imatrix argument to Anova() (which takes alist of matrices) or the P argument to linearHypothesis() (which takes a matrix). We illustratetrivially with a univariate ANOVA for the first response variable, Sepal.Length, extracted from themultivariate linear model for all four responses:

Anova(mod.iris, imatrix=list(Sepal.Length=matrix(c(1, 0, 0, 0))))

Type II Repeated Measures MANOVA Tests: Pillai test statistic


Sepal.Length 1 0.992 19327 1 147 <2e-16

Species:Sepal.Length 2 0.619 119 2 147 <2e-16

The univariate ANOVA for sepal length by species appears in the second line of the MANOVA tableproduced by Anova(). Similarly, using linearHypothesis(),

linearHypothesis(mod.iris, c("Speciesversicolor = 0", "Speciesvirginica = 0"),

P=matrix(c(1, 0, 0, 0))) # equivalent

Response transformation matrix:

[,1]

Sepal.Length 1

13

Sepal.Width 0

Petal.Length 0

Petal.Width 0


[,1]

[1,] 63.212


[,1]

[1,] 38.956

Multivariate Tests:


Pillai 2 0.61871 119.27 2 147 < 2.2e-16

Wilks 2 0.38129 119.27 2 147 < 2.2e-16


Roy 2 1.62265 119.27 2 147 < 2.2e-16

In this case, the P matrix is a single column picking out the first response. Finally, we verify thatwe get the same F -test from a univariate ANOVA for Sepal.Length:

Anova(lm(Sepal.Length ~ Species, data=iris))

Anova Table (Type II tests)

Response: Sepal.Length

Sum Sq Df F value Pr(>F)

Species 63.2 2 119 <2e-16

Residuals 39.0 147

Contrasts of the responses occur more naturally in the context of repeated-measures data, whichwe discuss in the following section.

3 Handling Repeated Measures

Repeated-measures data arise when multivariate responses represent the same individuals measuredon a response variable (or variables) on different occasions or under different circumstances. Theremay be a more or less complex design on the repeated measures. The simplest case is that of asingle repeated-measures or within-subjects factor, where the former term often is applied to datacollected over time and the latter when the responses represent different experimental conditions ortreatments. There may, however, be two or more within-subjects factors, as is the case, for example,when each subject is observed under different conditions on each of several occasions. The term“repeated measures” and “within-subjects factors” are common in disciplines, such as psychology,where the units of observation are individuals, but these designs are essentially the same as so-called“split-plot” designs in agriculture, where plots of land are each divided into sub-plots, which aresubjected to different experimental treatments, such as differing varieties of a crop or differing levelsof fertilizer.

Repeated-measures designs can be handled in R with the standard anova() function, as describedby Dalgaard (2007), but it is simpler to get common tests from the Anova() and linearHypothe-

sis() functions in the car package, as we explain in this section. The general procedure is first tofit a multivariate linear models with all of the repeated measures as responses; then an artificial data

14

frame is created in which each of the repeated measures is a row and in which the columns repre-sent the repeated-measures factor or factors; finally, the Anova() or linearHypothesis() functionis called, using the idata and idesign arguments (and optionally the icontrasts argument)—oralternatively the imatrix argument to Anova() or P argument to linearHypothesis()—to specifythe intra-subject design.

To illustrate, we employ contrived data reported by O’Brien and Kaiser (1985), in what they(justifiably) bill as “an extensive primer” for the MANOVA approach to repeated-measures designs.The data set OBrienKaiser is provided by the carData package:

some(OBrienKaiser)

treatment gender pre.1 pre.2 pre.3 pre.4 pre.5 post.1 post.2 post.3 post.4 post.5

2 control M 4 4 5 3 4 2 2 3 5 3

4 control F 5 4 7 5 4 2 2 3 5 3

5 control F 3 4 6 4 3 6 7 8 6 3

6 A M 7 8 7 9 9 9 9 10 8 9

7 A M 5 5 6 4 5 7 7 8 10 8

11 B M 3 3 4 2 3 5 4 7 5 4

12 B M 6 7 8 6 3 9 10 11 9 6

13 B F 5 5 6 8 6 4 6 6 8 6

14 B F 2 2 3 1 2 5 6 7 5 2

16 B F 4 5 7 5 4 7 7 8 6 7

fup.1 fup.2 fup.3 fup.4 fup.5

2 4 5 6 4 1

4 4 4 5 3 4

5 4 3 6 4 3

6 9 10 11 9 6

7 8 9 11 9 8

11 5 6 8 6 5

12 8 7 10 8 7

13 7 7 8 10 8

14 6 7 8 6 3

16 7 8 10 8 7

contrasts(OBrienKaiser$treatment)

[,1] [,2]

control -2 0

A 1 -1

B 1 1

contrasts(OBrienKaiser$gender)

[,1]

F 1

M -1

xtabs(~ treatment + gender, data=OBrienKaiser)

gender

treatment F M

control 2 3

A 2 2

B 4 3

15

There are two between-subjects factors in the O’Brien-Kaiser data: gender, with levels F and M; andtreatment, with levels A, B, and control. Both of these variables have predefined contrasts, with−1, 1 coding for gender and custom contrasts for treatment. In the latter case, the first contrastis for the control group versus the average of the experimental groups, and the second contrastis for treatment A versus treatment B. The frequency table for treatment by sex reveals that thedata are mildly unbalanced. We will imagine that the treatments A and B represent different inno-vative methods of teaching reading to learning-disabled students, and that the control treatmentrepresents a standard teaching method.

The 15 response variables in the data set represent two crossed within-subjects factors: phase,with three levels for the pretest, post-test, and follow-up phases of the study; and hour, representingfive successive hours, at which measurements of reading-comprehension are taken within each phase.We define the “data” for the within-subjects design as follows:

phase <- factor(rep(c("pretest", "posttest", "followup"), c(5, 5, 5)),

levels=c("pretest", "posttest", "followup"))

hour <- ordered(rep(1:5, 3))

idata <- data.frame(phase, hour)

idata

phase hour

1 pretest 1

2 pretest 2

3 pretest 3

4 pretest 4

5 pretest 5

6 posttest 1

7 posttest 2

8 posttest 3

9 posttest 4

10 posttest 5

11 followup 1

12 followup 2

13 followup 3

14 followup 4

15 followup 5

We begin by reshaping the data set from “wide” to “long” format to facilitate graphing the data;we will eventually use the original wide version of the data set for repeated-measures analysis.

OBrien.long <- reshape(OBrienKaiser,

varying=c("pre.1", "pre.2", "pre.3", "pre.4", "pre.5",

"post.1", "post.2", "post.3", "post.4", "post.5",

"fup.1", "fup.2", "fup.3", "fup.4", "fup.5"),

v.names="score",

timevar="phase.hour", direction="long")

OBrien.long$phase <- ordered(

c("pre", "post", "fup")[1 + ((OBrien.long$phase.hour - 1) %/% 5)],

levels=c("pre", "post", "fup"))

OBrien.long$hour <- ordered(1 + ((OBrien.long$phase.hour - 1) %% 5))

dim(OBrien.long)

[1] 240 7

16

head(OBrien.long, 25) # first 25 rows

treatment gender phase.hour score id phase hour

1.1 control M 1 1 1 pre 1



4.1 control F 1 5 4 pre 1


6.1 A M 1 7 6 pre 1

7.1 A M 1 5 7 pre 1

8.1 A F 1 2 8 pre 1

9.1 A F 1 3 9 pre 1

10.1 B M 1 4 10 pre 1

11.1 B M 1 3 11 pre 1

12.1 B M 1 6 12 pre 1

13.1 B F 1 5 13 pre 1

14.1 B F 1 2 14 pre 1

15.1 B F 1 2 15 pre 1

16.1 B F 1 4 16 pre 1






6.2 A M 2 8 6 pre 2

7.2 A M 2 5 7 pre 2

8.2 A F 2 3 8 pre 2

9.2 A F 2 3 9 pre 2

We then compute mean reading scores for combinations of gender, treatment, phase, and hour:

Means <- as.data.frame(ftable(with(OBrien.long,

tapply(score,

list(treatment=treatment, gender=gender, phase=phase, hour=hour),

mean))))

names(Means)[5] <- "score"

dim(Means)

[1] 90 5

head(Means, 25) # first 25 means

treatment gender phase hour score

1 control F pre 1 4.0000

2 A F pre 1 2.5000

3 B F pre 1 3.2500

4 control M pre 1 3.3333

5 A M pre 1 6.0000

6 B M pre 1 4.3333

7 control F post 1 4.0000

8 A F post 1 3.0000

9 B F post 1 5.5000

10 control M post 1 3.0000

17

11 A M post 1 8.0000

12 B M post 1 6.6667

13 control F fup 1 4.0000

14 A F fup 1 5.5000

15 B F fup 1 6.7500

16 control M fup 1 4.3333

17 A M fup 1 8.5000

18 B M fup 1 7.0000

19 control F pre 2 4.0000

20 A F pre 2 3.0000

21 B F pre 2 3.5000

22 control M pre 2 4.0000

23 A M pre 2 6.5000

24 B M pre 2 4.6667

25 control F post 2 4.5000

Finally, we employ the xyplot function in the lattice package to graph the means:4

library(lattice)

xyplot(score ~ hour | phase + treatment, groups=gender, type="b",

strip=function(...) strip.default(strip.names=c(TRUE, TRUE), ...),

lty=1:2, pch=c(15, 1), col=1:2, cex=1.25,

ylab="Mean Reading Score", data=Means,

key=list(title="Gender", cex.title=1,

text=list(c("Female", "Male")), lines=list(lty=1:2, col=1:2),

points=list(pch=c(15, 1), col=1:2, cex=1.25)))

The resulting graph is shown in Figure 4. It appears as if reading improves across phases in thetwo experimental treatments but not in the control group (suggesting a possible treatment-by-phaseinteraction); that there is a possibly quadratic relationship of reading to hour within each phase,with an initial rise and then decline, perhaps representing fatigue (suggesting an hour main effect);and that males and females respond similarly to the control and B treatment groups, but thatmales do better than females in the A treatment group (suggesting a possible gender-by-treatmentinteraction).

We next fit a multivariate linear model to the data, treating the repeated measures as responses,and with the between-subject factors treatment and gender (and their interaction) appearing onthe right-hand side of the model formula:

mod.ok <- lm(cbind(pre.1, pre.2, pre.3, pre.4, pre.5,

post.1, post.2, post.3, post.4, post.5,

fup.1, fup.2, fup.3, fup.4, fup.5) ~ treatment*gender,

data=OBrienKaiser)

mod.ok

Call:

lm(formula = cbind(pre.1, pre.2, pre.3, pre.4, pre.5, post.1,

post.2, post.3, post.4, post.5, fup.1, fup.2, fup.3, fup.4,

fup.5) ~ treatment * gender, data = OBrienKaiser)

Coefficients:

4Lattice graphics are described in Section 9.3.1 of the R Companion, and in more detail in Sarkar (2008).

18

hour

Mea

n R

eadi

ng S

core

4

6

8

10

1 2 3 4 5

: phase pre : treatment control

: phase post : treatment control

1 2 3 4 5

: phase fup : treatment control

: phase pre : treatment A

: phase post : treatment A

4

6

8

10

: phase fup : treatment A

4

6

8

10

: phase pre : treatment B

1 2 3 4 5

: phase post : treatment B

: phase fup : treatment B

GenderFemaleMale

Figure 4: Mean reading score by gender, treatment, phase, and hour, for the O’Brien-Kaiser data.

19

pre.1 pre.2 pre.3 pre.4 pre.5 post.1

(Intercept) 3.90e+00 4.28e+00 5.43e+00 4.61e+00 4.14e+00 5.03e+00

treatment1 1.18e-01 1.39e-01 -7.64e-02 1.81e-01 1.94e-01 7.64e-01

treatment2 -2.29e-01 -3.33e-01 -1.46e-01 -7.08e-01 -6.67e-01 2.92e-01

gender1 -6.53e-01 -7.78e-01 -1.81e-01 -1.11e-01 -6.39e-01 -8.61e-01

treatment1:gender1 -4.93e-01 -3.89e-01 -5.49e-01 -1.81e-01 -1.94e-01 -6.81e-01

treatment2:gender1 6.04e-01 5.83e-01 2.71e-01 7.08e-01 1.17e+00 9.58e-01

post.2 post.3 post.4 post.5 fup.1 fup.2

(Intercept) 5.54e+00 6.92e+00 6.36e+00 4.83e+00 6.01e+00 6.15e+00

treatment1 8.96e-01 8.33e-01 7.22e-01 9.17e-01 9.24e-01 1.03e+00

treatment2 1.87e-01 -2.50e-01 8.33e-02 8.70e-18 -6.25e-02 -6.25e-02

gender1 -4.58e-01 -4.17e-01 -5.28e-01 -1.00e+00 -5.97e-01 -9.03e-01

treatment1:gender1 -6.04e-01 -3.33e-01 -5.56e-01 -5.00e-01 -2.15e-01 -1.60e-01

treatment2:gender1 6.88e-01 2.50e-01 9.17e-01 1.25e+00 6.88e-01 1.19e+00

fup.3 fup.4 fup.5

(Intercept) 7.78e+00 6.17e+00 5.35e+00

treatment1 1.10e+00 9.58e-01 8.82e-01

treatment2 -1.25e-01 1.25e-01 2.29e-01

gender1 -7.78e-01 -8.33e-01 -4.31e-01

treatment1:gender1 -3.47e-01 -4.17e-02 -1.74e-01

treatment2:gender1 8.75e-01 1.12e+00 3.96e-01

We then compute the repeated-measures MANOVA using the Anova() function in the followingmanner:

(av.ok <- Anova(mod.ok, idata=idata, idesign=~phase*hour, type=3))

Type III Repeated Measures MANOVA Tests: Pillai test statistic


(Intercept) 1 0.967 296.4 1 10 9.2e-09

treatment 2 0.441 3.9 2 10 0.05471

gender 1 0.268 3.7 1 10 0.08480

treatment:gender 2 0.364 2.9 2 10 0.10447

phase 1 0.814 19.6 2 9 0.00052

treatment:phase 2 0.696 2.7 4 20 0.06211

gender:phase 1 0.066 0.3 2 9 0.73497

treatment:gender:phase 2 0.311 0.9 4 20 0.47215

hour 1 0.933 24.3 4 7 0.00033

treatment:hour 2 0.316 0.4 8 16 0.91833

gender:hour 1 0.339 0.9 4 7 0.51298

treatment:gender:hour 2 0.570 0.8 8 16 0.61319

phase:hour 1 0.560 0.5 8 3 0.82027

treatment:phase:hour 2 0.662 0.2 16 8 0.99155

gender:phase:hour 1 0.712 0.9 8 3 0.58949

treatment:gender:phase:hour 2 0.793 0.3 16 8 0.97237

� Following O’Brien and Kaiser (1985), we report type-III tests, by specifying the argumenttype=3. Although, as in univariate models, we generally prefer type-II tests (see Section 5.3.4of the R Companion), we wanted to preserve comparability with the original source. Type-IIItests are computed correctly because the contrasts employed for treatment and gender, andhence their interaction, are orthogonal in the row-basis of the between-subjects design. Weinvite the reader to compare these results with the default type-II tests.

20

� When, as here, the idata and idesign arguments are specified, Anova() automatically con-structs orthogonal contrasts for different terms in the within-subjects design, using contr.sum()

for a factor such as phase and contr.poly() (orthogonal polynomial contrasts) for an orderedfactor such as hour. Alternatively, the user can assign contrasts to the columns of the intra-subject data, either directly or via the icontrasts argument to Anova(). In any event,Anova() checks that the within-subjects contrast coding for different terms is orthogonal andreports an error if it is not.

� By default, Pillai’s test statistic is displayed; we invite the reader to examine the other threemultivariate test statistics.

� The results show that the anticipated hour effect has a small p-value, but the treatment ×phase and treatment × gender interactions have p-values that exceed 0.05. There is, however,a small p-values for the phase main effect. Of course, we should not over-interpret these results,partly because the data set is small and partly because it is contrived.

3.1 Univariate ANOVA for repeated measures

A traditional univariate approach to repeated-measures (or split-plot) designs (see, e.g., Winer, 1971,Chap. 7) computes an analysis of variance employing a “mixed-effects” models in which subjectsgenerate random effects. This approach makes stronger assumptions about the structure of thedata than the MANOVA approach described above, in particular stipulating that the covariancematrices for the repeated measures transformed by the within-subjects design (within combinationsof between-subjects factors) are spherical—that is, the transformed repeated measures for eachwithin-subjects test are uncorrelated and have the same variance, and this variance is constantacross cells of the between-subjects design. A sufficient (but not necessary) condition for sphericityof the errors is that the covariance matrix Σ of the repeated measures is compound-symmetric,with equal diagonal entries (representing constant variance for the repeated measures) and equaloff-diagonal elements (implying, together with constant variance, that the repeated measures havea constant correlation).

By default, when an intra-subject design is specified, summarizing the object produced byAnova() reports both MANOVA and univariate tests. Along with the traditional univariate tests,the summary reports tests for sphericity (Mauchly, 1940) and two corrections for non-sphericity ofthe univariate test statistics for within-subjects terms: the Greenhouse-Geiser correction (Green-house and Geisser, 1959) and the Huynh-Feldt correction (Huynh and Feldt, 1976). We illustratefor the O’Brien-Kaiser data, suppressing the multivariate tests:

summary(av.ok, multivariate=FALSE)

Univariate Type III Repeated-Measures ANOVA Assuming Sphericity

Sum Sq num Df Error SS den Df F value Pr(>F)

(Intercept) 6759 1 228.1 10 296.39 9.2e-09

treatment 180 2 228.1 10 3.94 0.0547

gender 83 1 228.1 10 3.66 0.0848

treatment:gender 130 2 228.1 10 2.86 0.1045

phase 130 2 80.3 20 16.13 6.7e-05

treatment:phase 78 4 80.3 20 4.85 0.0067

gender:phase 2 2 80.3 20 0.28 0.7566

treatment:gender:phase 10 4 80.3 20 0.64 0.6424

hour 104 4 62.5 40 16.69 4.0e-08

treatment:hour 1 8 62.5 40 0.09 0.9992

21

gender:hour 3 4 62.5 40 0.45 0.7716

treatment:gender:hour 8 8 62.5 40 0.62 0.7555

phase:hour 11 8 96.2 80 1.18 0.3216

treatment:phase:hour 7 16 96.2 80 0.35 0.9901

gender:phase:hour 9 8 96.2 80 0.93 0.4956

treatment:gender:phase:hour 14 16 96.2 80 0.74 0.7496

Mauchly Tests for Sphericity

Test statistic p-value

phase 0.749 0.273

treatment:phase 0.749 0.273

gender:phase 0.749 0.273

treatment:gender:phase 0.749 0.273

hour 0.066 0.008

treatment:hour 0.066 0.008

gender:hour 0.066 0.008

treatment:gender:hour 0.066 0.008

phase:hour 0.005 0.449

treatment:phase:hour 0.005 0.449

gender:phase:hour 0.005 0.449

treatment:gender:phase:hour 0.005 0.449

Greenhouse-Geisser and Huynh-Feldt Corrections

for Departure from Sphericity

GG eps Pr(>F[GG])

phase 0.80 0.00028

treatment:phase 0.80 0.01269

gender:phase 0.80 0.70896

treatment:gender:phase 0.80 0.61162

hour 0.46 9.8e-05

treatment:hour 0.46 0.97862

gender:hour 0.46 0.62843

treatment:gender:hour 0.46 0.64136

phase:hour 0.45 0.33452

treatment:phase:hour 0.45 0.93037

gender:phase:hour 0.45 0.44908

treatment:gender:phase:hour 0.45 0.64634

HF eps Pr(>F[HF])

phase 0.92786 1.1247e-04

treatment:phase 0.92786 8.4388e-03

gender:phase 0.92786 7.4086e-01

treatment:gender:phase 0.92786 6.3200e-01

hour 0.55928 2.3009e-05

treatment:hour 0.55928 9.8866e-01

gender:hour 0.55928 6.6455e-01

treatment:gender:hour 0.55928 6.6930e-01

22

phase:hour 0.73306 3.2966e-01

treatment:phase:hour 0.73306 9.7523e-01

gender:phase:hour 0.73306 4.7803e-01

treatment:gender:phase:hour 0.73306 7.0801e-01

The non-sphericity tests have small p-values for F -tests involving hour; the results for the univariateANOVA are not terribly different from those of the MANOVA reported above, except that now thetreatment × phase interaction is associated with a p-value smaller than 0.05.

3.2 Using linearHypothesis() with repeated-measures designs

As for simpler multivariate linear models (discussed in Section 2), the linearHypothesis() functioncan be used to test more focused hypotheses about the parameters of repeated-measures models,including for within-subjects terms.

As a preliminary example, to reproduce the test for the main effect of hour, we can use theidata, idesign, and iterm arguments in a call to linearHypothesis():

linearHypothesis(mod.ok, "(Intercept) = 0", idata=idata,

idesign=~phase*hour, iterms="hour") # test hour main effect


hour.L hour.Q hour.C hour^4

pre.1 -0.63246 0.53452 -3.1623e-01 0.11952

pre.2 -0.31623 -0.26726 6.3246e-01 -0.47809

pre.3 0.00000 -0.53452 -4.0960e-16 0.71714

pre.4 0.31623 -0.26726 -6.3246e-01 -0.47809

pre.5 0.63246 0.53452 3.1623e-01 0.11952

post.1 -0.63246 0.53452 -3.1623e-01 0.11952

post.2 -0.31623 -0.26726 6.3246e-01 -0.47809

post.3 0.00000 -0.53452 -4.0960e-16 0.71714

post.4 0.31623 -0.26726 -6.3246e-01 -0.47809

post.5 0.63246 0.53452 3.1623e-01 0.11952

fup.1 -0.63246 0.53452 -3.1623e-01 0.11952

fup.2 -0.31623 -0.26726 6.3246e-01 -0.47809

fup.3 0.00000 -0.53452 -4.0960e-16 0.71714

fup.4 0.31623 -0.26726 -6.3246e-01 -0.47809

fup.5 0.63246 0.53452 3.1623e-01 0.11952



hour.L 0.010345 1.5562 0.36724 -0.82435

hour.Q 1.556250 234.1182 55.24686 -124.01365

hour.C 0.367241 55.2469 13.03707 -29.26455

hour^4 -0.824354 -124.0137 -29.26455 65.69068



hour.L 89.7333 49.6106 -9.7167 -25.418

hour.Q 49.6106 46.6429 1.3522 -17.409

hour.C -9.7167 1.3522 21.8083 16.111

hour^4 -25.4181 -17.4094 16.1107 29.315

23

Multivariate Tests:


Pillai 1 0.9329 24.315 4 7 0.0003345

Wilks 1 0.0671 24.315 4 7 0.0003345

Hotelling-Lawley 1 13.8944 24.315 4 7 0.0003345

Roy 1 13.8944 24.315 4 7 0.0003345

Because hour is a within-subjects factor, we test its main effect as the regression intercept in thebetween-subjects model, using a response-transformation matrix for the hour contrasts.

Alternatively and equivalently, we can generate the response-transformation matrix P for thehypothesis directly:

(Hour <- model.matrix(~ hour, data=idata))

(Intercept) hour.L hour.Q hour.C hour^4

1 1 -0.63246 0.53452 -3.1623e-01 0.11952

2 1 -0.31623 -0.26726 6.3246e-01 -0.47809

3 1 0.00000 -0.53452 -4.0960e-16 0.71714

4 1 0.31623 -0.26726 -6.3246e-01 -0.47809

5 1 0.63246 0.53452 3.1623e-01 0.11952

6 1 -0.63246 0.53452 -3.1623e-01 0.11952

7 1 -0.31623 -0.26726 6.3246e-01 -0.47809

8 1 0.00000 -0.53452 -4.0960e-16 0.71714

9 1 0.31623 -0.26726 -6.3246e-01 -0.47809

10 1 0.63246 0.53452 3.1623e-01 0.11952

11 1 -0.63246 0.53452 -3.1623e-01 0.11952

12 1 -0.31623 -0.26726 6.3246e-01 -0.47809

13 1 0.00000 -0.53452 -4.0960e-16 0.71714

14 1 0.31623 -0.26726 -6.3246e-01 -0.47809

15 1 0.63246 0.53452 3.1623e-01 0.11952

attr(,"assign")

[1] 0 1 1 1 1

attr(,"contrasts")

attr(,"contrasts")$hour

[1] "contr.poly"

linearHypothesis(mod.ok, "(Intercept) = 0",

P=Hour[ , c(2:5)]) # test hour main effect (equivalent)



pre.1 -0.63246 0.53452 -3.1623e-01 0.11952

pre.2 -0.31623 -0.26726 6.3246e-01 -0.47809

pre.3 0.00000 -0.53452 -4.0960e-16 0.71714

pre.4 0.31623 -0.26726 -6.3246e-01 -0.47809

pre.5 0.63246 0.53452 3.1623e-01 0.11952

post.1 -0.63246 0.53452 -3.1623e-01 0.11952

post.2 -0.31623 -0.26726 6.3246e-01 -0.47809

post.3 0.00000 -0.53452 -4.0960e-16 0.71714

post.4 0.31623 -0.26726 -6.3246e-01 -0.47809

24

post.5 0.63246 0.53452 3.1623e-01 0.11952

fup.1 -0.63246 0.53452 -3.1623e-01 0.11952

fup.2 -0.31623 -0.26726 6.3246e-01 -0.47809

fup.3 0.00000 -0.53452 -4.0960e-16 0.71714

fup.4 0.31623 -0.26726 -6.3246e-01 -0.47809

fup.5 0.63246 0.53452 3.1623e-01 0.11952



hour.L 0.010345 1.5562 0.36724 -0.82435

hour.Q 1.556250 234.1182 55.24686 -124.01365

hour.C 0.367241 55.2469 13.03707 -29.26455

hour^4 -0.824354 -124.0137 -29.26455 65.69068



hour.L 89.7333 49.6106 -9.7167 -25.418

hour.Q 49.6106 46.6429 1.3522 -17.409

hour.C -9.7167 1.3522 21.8083 16.111

hour^4 -25.4181 -17.4094 16.1107 29.315

Multivariate Tests:


Pillai 1 0.9329 24.315 4 7 0.0003345

Wilks 1 0.0671 24.315 4 7 0.0003345


Roy 1 13.8944 24.315 4 7 0.0003345

As mentioned, this test simply duplicates part of the output from Anova(), but suppose that wewant to test the individual polynomial components of the hour main effect:

linearHypothesis(mod.ok, "(Intercept) = 0", P=Hour[ , 2, drop=FALSE]) # linear


hour.L

pre.1 -0.63246

pre.2 -0.31623

pre.3 0.00000

pre.4 0.31623

pre.5 0.63246

post.1 -0.63246

post.2 -0.31623

post.3 0.00000

post.4 0.31623

post.5 0.63246

fup.1 -0.63246

fup.2 -0.31623

fup.3 0.00000

fup.4 0.31623

fup.5 0.63246

25


hour.L

hour.L 0.010345


hour.L

hour.L 89.733

Multivariate Tests:


Pillai 1 0.00012 0.0011528 1 10 0.9736

Wilks 1 0.99988 0.0011528 1 10 0.9736


Roy 1 0.00012 0.0011528 1 10 0.9736

linearHypothesis(mod.ok, "(Intercept) = 0", P=Hour[ , 3, drop=FALSE]) # quadratic


hour.Q

pre.1 0.53452

pre.2 -0.26726

pre.3 -0.53452

pre.4 -0.26726

pre.5 0.53452

post.1 0.53452

post.2 -0.26726

post.3 -0.53452

post.4 -0.26726

post.5 0.53452

fup.1 0.53452

fup.2 -0.26726

fup.3 -0.53452

fup.4 -0.26726

fup.5 0.53452


hour.Q

hour.Q 234.12


hour.Q

hour.Q 46.643

Multivariate Tests:


Pillai 1 0.8339 50.194 1 10 3.356e-05

Wilks 1 0.1661 50.194 1 10 3.356e-05

Hotelling-Lawley 1 5.0194 50.194 1 10 3.356e-05

Roy 1 5.0194 50.194 1 10 3.356e-05

linearHypothesis(mod.ok, "(Intercept) = 0", P=Hour[ , 4, drop=FALSE]) # cubic

26


hour.C

pre.1 -3.1623e-01

pre.2 6.3246e-01

pre.3 -4.0960e-16

pre.4 -6.3246e-01

pre.5 3.1623e-01

post.1 -3.1623e-01

post.2 6.3246e-01

post.3 -4.0960e-16

post.4 -6.3246e-01

post.5 3.1623e-01

fup.1 -3.1623e-01

fup.2 6.3246e-01

fup.3 -4.0960e-16

fup.4 -6.3246e-01

fup.5 3.1623e-01


hour.C

hour.C 13.037


hour.C

hour.C 21.808

Multivariate Tests:


Pillai 1 0.37414 5.978 1 10 0.03455

Wilks 1 0.62586 5.978 1 10 0.03455


Roy 1 0.59780 5.978 1 10 0.03455

linearHypothesis(mod.ok, "(Intercept) = 0", P=Hour[ , 5, drop=FALSE]) # quartic


hour^4

pre.1 0.11952

pre.2 -0.47809

pre.3 0.71714

pre.4 -0.47809

pre.5 0.11952

post.1 0.11952

post.2 -0.47809

post.3 0.71714

post.4 -0.47809

post.5 0.11952

fup.1 0.11952

fup.2 -0.47809

fup.3 0.71714

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fup.4 -0.47809

fup.5 0.11952


hour^4

hour^4 65.691


hour^4

hour^4 29.315

Multivariate Tests:


Pillai 1 0.69144 22.408 1 10 0.0007997

Wilks 1 0.30856 22.408 1 10 0.0007997


Roy 1 2.24082 22.408 1 10 0.0007997

linearHypothesis(mod.ok, "(Intercept) = 0", P=Hour[ , c(2, 4:5)]) # all non-quadratic


hour.L hour.C hour^4

pre.1 -0.63246 -3.1623e-01 0.11952

pre.2 -0.31623 6.3246e-01 -0.47809

pre.3 0.00000 -4.0960e-16 0.71714

pre.4 0.31623 -6.3246e-01 -0.47809

pre.5 0.63246 3.1623e-01 0.11952

post.1 -0.63246 -3.1623e-01 0.11952

post.2 -0.31623 6.3246e-01 -0.47809

post.3 0.00000 -4.0960e-16 0.71714

post.4 0.31623 -6.3246e-01 -0.47809

post.5 0.63246 3.1623e-01 0.11952

fup.1 -0.63246 -3.1623e-01 0.11952

fup.2 -0.31623 6.3246e-01 -0.47809

fup.3 0.00000 -4.0960e-16 0.71714

fup.4 0.31623 -6.3246e-01 -0.47809

fup.5 0.63246 3.1623e-01 0.11952



hour.L 0.010345 0.36724 -0.82435

hour.C 0.367241 13.03707 -29.26455

hour^4 -0.824354 -29.26455 65.69068



hour.L 89.7333 -9.7167 -25.418

hour.C -9.7167 21.8083 16.111

hour^4 -25.4181 16.1107 29.315

Multivariate Tests:

28


Pillai 1 0.8963 23.05 3 8 0.0002724

Wilks 1 0.1037 23.05 3 8 0.0002724


Roy 1 8.6439 23.05 3 8 0.0002724

The hour main effect is more complex, therefore, than a simple quadratic trend.

4 Complementary Reading and References

The material in the first section of this appendix is based on Fox (2016, Sec. 9.5).There are many texts that treat MANOVA and multivariate linear models: The theory is pre-

sented in Rao (1973); more generally accessible treatments include Hand and Taylor (1987) andMorrison (2005). A good, brief introduction to the MANOVA approach to repeated-measures maybe found in O’Brien and Kaiser (1985). As mentioned, Winer (1971, Chap. 7) presents the traditionalunivariate approach to repeated-measures.

References

Anderson, E. (1935). The irises of the Gaspe Peninsula. Bulletin of the American Iris Society,59:2–5.

Dalgaard, P. (2007). New functions for multivariate analysis. R News, 7(2):2–7.

Fisher, R. A. (1936). The use of multiple measurements in taxonomic problems. Annals of Eugenics,7, Part II:179–188.

Fox, J. (2016). Applied Regression Analysis and Generalized Linear Models. Sage, Thousand OaksCA, third edition.

Fox, J., Friendly, M., and Monette, G. (2009). Visualizing hypothesis tests in multivariate linearmodels: The heplots package for R. Computational Statistics, 24:233–246.

Fox, J. and Weisberg, S. (2019). An R Companion to Applied Regression. Sage, Thousand Oaks,CA, third edition.

Friendly, M. (2007). HE plots for multivariate linear models. Journal of Computational and GraphicalStatistics, 16:421–444.

Friendly, M. (2010). HE plots for repeated measures designs. Journal of Statistical Software, 37(4):1–40.

Greenhouse, S. W. and Geisser, S. (1959). On methods in the analysis of profile data. Psychometrika,24:95–112.

Hand, D. J. and Taylor, C. C. (1987). Multivariate Analysis of Variance and Repeated Measures: APractical Approach for Behavioural Scientists. Chapman and Hall, London.

Huynh, H. and Feldt, L. S. (1976). Estimation of the Box correction for degrees of freedom fromsample data in randomized block and split-plot designs. Journal of Educational Statistics, 1:69–82.

Mauchly, J. W. (1940). Significance test for sphericity of a normal n-variate distribution. The Annalsof Mathematical Statistics, 11:204–209.

Morrison, D. F. (2005). Multivariate Statistical Methods. Duxbury, Belmont CA, 4th edition.

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O’Brien, R. G. and Kaiser, M. K. (1985). MANOVA method for analyzing repeated measuresdesigns: An extensive primer. Psychological Bulletin, 97:316–333.

Rao, C. R. (1973). Linear Statistical Inference and Its Applications. Wiley, New York, secondedition.

Sarkar, D. (2008). Lattice: Multivariate Data Visualization with R. Springer, New York.

Winer, B. J. (1971). Statistical Principles in Experimental Design. McGraw-Hill, New York, secondedition.

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